On a single Argand diagram sketch the loci |z| = 5 and |z-5|=|z|. Hence determine the complex numbers represented by points common to both loci, giving each answer in the exponential form.
I know that we have to draw a circle with centre at the origin and radius 5 for |z|=5 ,and then its circumference will be the required figure .For the second one,I am not sure what to do.How then should I answer the second part of the question?
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Consider the second constraint.
Let $A=(5,0), B=(0,0)$ and $M$ be the generic point of the locus.
$$|z-5|=|z-0| \iff dist(M,A)=dist(M,B)$$
The set of such points $M$ is the perpendicular bissector of line segment $[AB]$.
Therefore, it is the line with cartesian equation $x=5/2$ and the Argand plane equation
$$z=x+iy=5/2+it \ \ \text{for any real} \ t \tag{1}$$
We want those points that are such that (first constraint) $|z|=5$, meaning that
$$|z|^2=25=25/4+t^2 \iff t_0=\pm(5/2)\sqrt{3}$$
The argument of such a $z$ is (by reference to (1)) can be computed (as the two solutions are in quadrants 1 and 4) by :
$$\tan^{-1}(\frac{y}{x})=\tan^{-1}\frac{t_0}{(5/2)}=\pm \tan^{-1}\sqrt{3}=\pm \frac{\pi}{3}$$
Final answer : $z=5e^{\pm i\tfrac{\pi}{3}}$
Write $|z-5|=|z|$ as $(z-5)(\bar z-5) = z\bar z$. Simplify, along with $|z|^2 =25$, to get,
$$z+\bar z = 5, \>\>\>\>\> z\bar z=25 \tag 1$$
Note that the first equation is also $Re(z) = \frac52$, which is the vertical line at $x=\frac52$ in the Argand diagram.
The points common to both $|z|=5$ and $|z-5|=|z|$ are the intersections between the two equations in (1), which can be combined as,
$$z^2-5z+25 = \frac{z^3+125}{z+5}=0$$
and yields two points given by $z^3=-125$, i.e.
$$z=5e^{\pm i \frac\pi3}$$